Continuous Time Signals (Part - I) – Fourier series

(a) Basics

1. Which of the following signals is/are periodic?

(a) 𝑠(𝑡) = cos 2𝑡 + cos 3𝑡 + cos 5𝑡

(b) 𝑠(𝑡) = exp(𝑗8 𝜋𝑡)

(c) 𝑠(𝑡) = exp(−7𝑡) sin 10𝜋𝑡

(d) 𝑠(𝑡) = cos 2𝑡 cos 4𝑡

[GATE 1992: 2 Marks]

Soln. (a) 𝑠(𝑡) = cos 2 𝑡 + cos 3 𝑡 + cos 5 𝑡

First term has 𝜔1 = 2

Second term has 𝜔2 = 3

Third term has 𝜔3 = 5

Note that ratio of any two frequencies equals p/q is rational where

p and q are integers.

Thus 𝑠(𝑡) is periodic

(b) 𝑠(𝑡) = exp(𝑗8𝜋𝑡)

= cos(8𝜋𝑡) + 𝑗 sin(8𝜋𝑡)

𝜔1

𝜔2=

8𝜋

8𝜋= 1 𝑠𝑜 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐

(c) 𝑠(𝑡) = exp(−7𝑡) . sin 10𝜋𝑡

= 𝑒−7𝑡 . sin 10𝜋𝑡

𝑁𝑜𝑡𝑒, 𝑇 =2𝜋

𝜔=

2𝜋

10𝜋=

1

5

Due to 𝑒−7𝑡 it is decaying function, so not periodic

(d) 𝑠(𝑡) = cos 2 𝑡 . cos 4 𝑡

Note, 2 cos𝐴 cos𝐵 = cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵)

𝑠𝑜, 𝑠(𝑡) =1

2[cos 2 𝑡 + cos 6 𝑡]

𝑁𝑜𝑡𝑒, 𝑝

𝑞=

𝜔1

𝜔2=

2

6=

1

3

Rational

So, (a) , (b) and (d) are periodic.

2. The power in the signal

𝑠(𝑡) = 8𝑐𝑜𝑠 (20𝜋𝑡 −𝜋

2) + 4𝑠𝑖𝑛(15𝜋𝑡) 𝑖𝑠

(a) 40

(b) 41

(c) 42

(d) 82

[GATE 2005: 1 Mark]

Soln. Time average of energy of a signal = Power of Signal

= lim𝑇→∞

1

𝑇∫ 𝑓2(𝑡)𝑑𝑡 = lim

𝑇→∞

1

𝑇∫ |𝑓(𝑡)|2𝑑𝑡

𝑇/2

−𝑇/2

𝑇/2

−𝑇/2

Signal power P is mean of the signal amplitude squared value of

f(t) . Rms value of signal = √𝑃

𝑆(𝑡) = 8 cos (20𝜋𝑡 −𝜋

2) + 4 sin(15𝜋𝑡)

= 8 sin(20𝜋𝑡) + 4 sin(15𝜋𝑡)

=82

2+

42

2= 32 + 8 = 40

Option (a)

3. If a signal f(t) has energy E, the energy of the signal f(2t) is equal to

(a) E

(b) E/2

(c) 2E

(d) 4E

[GATE 2001: 1 Mark]

Soln. Energy of a signal is given by

𝐸 = ∫[𝑓(𝑡)]2𝑑𝑡

∞

−∞

Energy of the signal f(2t) is

𝐸𝑠 = ∫[𝑓(𝑡)]2 𝑑𝑡

∞

−∞

Let 2𝑡 = 𝑝 𝑜𝑟, 𝑑𝑡 =𝑑𝑝

2

= ∫[𝑓(𝑡)]2 𝑑𝑝

2

∞

−∞

𝐸𝑠 =𝐸

2

Option (b)

4. For a periodic signal

𝑣(𝑡) = 30 sin 100𝑡 + 10 cos 300𝑡 + 6 sin(500𝑡 +𝜋

4), the fundamental

frequency in rad/s is

(a) 100

(b) 300

(c) 500

(d) 1500

[GATE 2013: 1 Mark]

Soln. First term has 𝜔1 = 100

Second term 𝜔2 = 300

Third term 𝜔3 = 500

𝜔1is the fundamental frequency

𝜔2is third harmonic

𝜔3is 5th harmonic

Option (a)

5. Consider the periodic square wave in the figure shown

x

1 2 3 4 t

-1

0

The ratio of the power in the 7th harmonic to the power in the 5th

harmonic for this waveform is closest in value to ------

[GATE 2014: 1 Mark]

Soln. For a periodic square wave nth harmonic component ∝1

𝑛

Thus the power in the nth harmonic component is ∝ 1𝑛2⁄

Ratio of power in 7th harmonic to 5th harmonic for the given wage form is

172⁄

152⁄

=25

29 ≅ 0.5

Answer 0.5

6. The waveform of a periodic signal 𝑥(𝑡) is shown in the figure.

-4 -3 2 3-2

-1

-3

3

1t4

A signal 𝑔(𝑡) is defined by𝑔(𝑡) = 𝑥 (𝑡−1

2). The average power of 𝑔(𝑡) is

_________ .

[GATE 2015: 1 Mark]

Soln. The equation for the given waveform can be written as

𝑥 = −3 𝑡

The period of the waveform is 3 (i.e. from -1 to +2)

𝐴𝑣. 𝑃𝑜𝑤𝑒𝑟 = 1

𝑇∫[𝑥(𝑡)]2 𝑑𝑡

=1

3[ ∫(−3 𝑡)2 𝑑𝑡 + ∫(−3 𝑡)2 𝑑𝑡 + ∫02. 𝑑𝑡

2

1

1

0

0

−1

]

=1

3[9.

𝑡3

3 |

0−1

+ 9 𝑡3

3 |10

+ 0]

=1

3[9

3. {0 − (−1)} +

9

3(1 − 0)]

=1

3[9

3+

9

3] =

1

3×

18

3= 2

Answer 2

7. The RMS value of a rectangular wave of period T, having a value of +V

for a duration 𝑇1 (< 𝑇) 𝑎𝑛𝑑 − 𝑉 for the duration 𝑇 − 𝑇1 = 𝑇2, equals

(a) V

(b) 𝑇1−𝑇2

𝑇𝑉

(c) 𝑉

√2

(d) 𝑇1

𝑇2𝑉

[GATE: 1995 1 Mark]

Soln.

-V

+V

Tt

The waveform can be drawn as per the given problem.

Period (𝑇) = 𝑇1 + 𝑇2

𝑅𝑀𝑆 𝑣𝑎𝑙𝑢𝑒 = √1

𝑇∫ 𝑥2(𝑡) 𝑑𝑡

𝑇

0

= √1

𝑇[∫ 𝑉2 𝑑𝑡 + ∫(−𝑉)2 𝑑𝑡

𝑇

𝑇1

𝑇1

0

]

= √1

𝑇[𝑉2. (𝑇1 − 0) + 𝑉2(𝑇 − 𝑇1)]

= √1

𝑇. 𝑉2 [ 𝑇1 + 𝑇 − 𝑇1] = √𝑉2 = 𝑉 𝑂𝑝𝑡𝑖𝑜𝑛 (𝑎)

(b) Fourier series

8. The trigonometric Fourier series of an even function of time does not

have

(a) the dc term

(b) cosine terms

(c) sine terms

(d) odd harmonic terms

[GATE 1996: 1 Mark]

Soln. For periodic even function, the trigonometric Fourier series does not

contain the sine terms (odd functions)

It has dc term and cosine terms of all harmonics.

Option (c)

9. The trigonometric Fourier series of a periodic time function can have

only

(a) cosine terms

(b) sine terms

(c) cosine and sine terms

(d) dc and cosine terms

[GATE 1998: 1 Mark]

Soln. The Fourier series of a periodic function 𝑥(𝑡) is given by the form

𝑥(𝑡) = ∑ 𝑎𝑛 cos 𝑛 𝜔0𝑡 + ∑ 𝑏𝑛 sin 𝑛 𝜔0 𝑡

∞

𝑛=1

∞

𝑛−0

Thus the series has cosine terms of all harmonics: 𝑛𝜔0, 𝑛 = 0,1,2 − − −

Where 0th harmonic = dc term (average or mean) = a0 and sine terms of

all harmonics: 𝑛𝜔0, 𝑛 = 1,2,− − − −.

10. The Fourier series of an odd periodic function, contains only

(a) odd harmonics

(b) even harmonics

(c) cosine terms

(d) sine terms

[GATE 1994: 1 Mark]

Soln. If periodic function is odd the dc term 𝑎0 = 0 and also cosine terms (even

symmetry)

It contains only sine terms

Option (d)

11. The Fourier series of a real periodic function has only

P. Cosine terms if it is even

Q. Sine terms if it is even

R. Cosine terms if it is odd

S. Sine terms if it odd

Which of the above statements are correct?

(a) P and S

(b) P and R

(c) Q and S

(d) Q and R

[GATE 2009: 1 Mark]

Soln. The Fourier series for a real periodic function has only cosine terms if it

is even and sine terms if it is odd

Option (a)

12. The trigonometric Fourier series of an even function does not have the

(a) dc term

(b) cosine terms

(c) sine terms

(d) odd harmonic terms

[GATE 2011: 1 Mark]

Soln. The trigonometric Fourier series of an even function has cosine terms

which are even functions.

It has dc term if its average value is finite and no dc term if average value

is zero

So it does not have sine terms

Option (c)

13. Which of the following cannot be the Fourier series expansion of a

periodic signals?

(a) 𝑥(𝑡) = 2 cos 𝑡 + 3 cos 3𝑡

(b) 𝑥(𝑡) = 2 cos 𝜋𝑡 + 7 cos 𝑡

(c) 𝑥(𝑡) = cos 𝑡 + 0.5

(d) 𝑥(𝑡)2 cos 1.5𝜋𝑡 + sin 3.5𝜋𝑡

[GATE 2002: 1 Mark]

Soln. (a) 𝑥(𝑡) = 2 cos 𝑡 + 3 cos 𝑡 is periodic signal with fundamental

frequency 𝜔0 = 1

(b) 𝑥(𝑡) = 2 cos 𝜋 𝑡 + 7 cos 𝑡 The frequency of first term 𝜔1 = 𝜋

frequency of 2nd term is 𝜔2 = 1

𝜔1

𝜔2=

𝜋

1 𝑖𝑠 𝑛𝑜𝑡 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟

So 𝑥(𝑡) is aperiodic or not periodic

(c) 𝑥(𝑡) = cos 𝑡 + 0.5 is a periodic function with 𝜔0 = 1

(d) 𝑥(𝑡) = 2 cos(1.5𝜋)𝑡 + sin(3.5𝜋)𝑡 first term has frequency 𝜔1 =

1.5𝜋 2nd term has frequency 𝜔2 = 3.5𝜋

𝜔1

𝜔2=

1.5𝜋

3.5𝜋=

1.5

3.5=

3 × 0.5

7 × 0.5=

3

7

So about ratio is rational number 𝑥(𝑡) is a periodic signal, with

fundamental frequency 𝜔0 = 0.5𝜋

Since function in (b) is non periodic. So does not satisfy Dirictilet

condition and cannot be expanded in Fourier series

14. Choose the function 𝑓(𝑡), −∞ < 𝑡 < ∞, for which a Fourier series

cannot be defined.

(a) 3 sin(25𝑡)

(b) 4 cos(20𝑡 + 3) + 2 sin(710𝑡)

(c) exp(−|𝑡|) sin(25𝑡)

(d) 1

[GATE 2005: 1 Mark]

Soln. Fourier series is defined for periodic function and constant

(a) 3 sin(25 𝑡) 𝑖𝑠 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝜔 = 25

(b) 4 cos(20 𝑡 + 3) + 2 sin(710 𝑡) sum of two periodic function is also

periodic function

(c) 𝑒−|𝑡| sin 25 𝑡 Due to decaying exponential decaying function it is not

periodic. So Fourier series cannot be defined for it.

(d) Constant, Fourier series exists.

Fourier series can’t be defined for option (c)

15. A periodic signal 𝑥(𝑡)𝑜𝑓 𝑝𝑒𝑟𝑖𝑜𝑑 𝑇0 is given by

𝑥(𝑡) = {1, |𝑡| < 𝑇1

0, 𝑇1 < |𝑡| <𝑇0

2

The dc component of 𝑥(𝑡) is

(a) 𝑇1

𝑇0

(b) 𝑇1

(2𝑇0)

(c) 2𝑇1

𝑇0

(d) 𝑇0

𝑇1

[GATE 1998: 1 Mark]

Soln.

-T

1

Given periodic signal can be drawn having period T0

Fourier series the function 𝑥(𝑡) can be written as

𝑥(𝑡) = 𝑎0 + ∑ 𝑎𝑛

∞

𝑛=1

cos 𝑛𝜔𝑜𝑡 + 𝑏𝑛 sin 𝑛𝜔0𝑡

Where dc component given by

𝑎0 =1

𝑇0∫ 𝑥(𝑡) 𝑑𝑡

∞

𝑇0

𝑎0 =1

𝑇∫ 𝑥(𝑡)𝑑𝑡

𝑇02⁄

−𝑇02⁄

=1

𝑇0

[

∫ 𝑥(𝑡)𝑑𝑡 + ∫ 𝑥(𝑡)𝑑𝑡 + ∫ 𝑥(𝑡)𝑑𝑡

𝑇02⁄

𝑇1

𝑇1

−𝑇1

−𝑇1

−𝑇02⁄ ]

=1

𝑇0

[0 + 2𝑇1 + 0]

=2𝑇1

𝑇0

Option (c)

16. The Fourier series representation of an impulse train denoted by

𝑠(𝑡) = ∑ 𝛿(𝑡 − 𝑛𝑇0)

∞

𝑛=−∞

𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦

(𝑎) (1

𝑇0) ∑ exp(−𝑗2𝜋𝑛 𝑡/𝑇0)

∞

𝑛=−∞

(𝑏) (1

𝑇0) ∑ exp(−𝑗𝜋𝑛 𝑡/𝑇0)

∞

𝑛=−∞

(𝑐) (1

𝑇0) ∑ exp(𝑗𝜋𝑛 𝑡/𝑇0)

∞

𝑛=−∞

(𝑑) (1

𝑇0) ∑ exp(𝑗2𝜋𝑛 𝑡/𝑇0)

∞

𝑛=−∞

Soln.

0

The given impulse train 𝑠(𝑡) with strength of each impulse as 1 is

aperiodic function with period T0

𝑠(𝑡) = ∑ 𝐶𝑛𝑒𝑗𝑛𝜔0𝑡

∞

𝑛=−∞

𝑤ℎ𝑒𝑟𝑒 𝜔0 =2𝑇

𝑇0

𝑤ℎ𝑒𝑟𝑒 𝐶𝑛 =1

𝑇0∫ 𝛿(𝑡)𝑒−𝑗𝑛𝜔0𝑡 𝑑𝑡

+𝑇02⁄

−𝑇02⁄

=1. 𝑒−𝑗𝑛𝜔0𝑡

𝑇0|𝑡 = 0 = 1

𝑇0⁄

17. The Fourier series expansion of a real periodic signal with fundamental

frequency f0 is given by

𝑔𝑝(𝑡) = ∑ 𝐶𝑛 𝑒𝑗 2𝜋𝑛 𝑓0 𝑡

∞

𝑛=−∞

It is given that 𝐶3 = 3 + 𝑗5 𝑡ℎ𝑒𝑛 𝐶−3 𝑖𝑠

(a) 5 + 3𝑗

(b) −3 − 𝑗5

(c) −5 + 3𝑗 (d) 3 −j5

[GATE 2003: 1 Mark]

Soln. Given 𝐶3 = 3 + 𝑗5

We know that for real periodic signal

𝐶−𝑘 = 𝐶𝑘∗

𝑆𝑜, 𝐶−3 = 𝐶3∗ = (3 − 𝑗5)